Notation: We write $\mathbb E_{\mathcal F} X$ for the conditional expectation $\mathbb E[X|\mathcal F]$ of a random variable $X$ with respect to a $\sigma$-algebra $\mathcal F$.
Let $X$ be an integrable random variable with associated $\sigma$-algebra $\mathcal X$, and let $\mathcal G, \mathcal H$ be sub $\sigma$-algebras of $\mathcal X$. It is known that the seemingly intuitive identity
$$\mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X = \mathbb E_{\mathcal H} \mathbb E_{\mathcal G} X = \mathbb E_{\mathcal G \cap \mathcal H} X$$
does not hold in general, see for example here for an elementary construction where all three of the values above are different.
However, I believe the following holds.
Question: Is it true that the sequence
$$\mathbb E_{\mathcal H} X, \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X, \mathbb E_{\mathcal H} \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X, \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X, \dots$$
converges almost surely to $ \mathbb E_{\mathcal G \cap \mathcal H} X$?
Comments: The desired result seems to be true if $X$ is a random variable taking finitely many values. Indeed, viewing the conditional expectation operator as a projection, the convergence is guaranteed by the following algorithm, popularised by von Neumann.